\u4eca\u5929\u6211\u4eec\u8ba8\u8bba\u751f\u6210\u51fd\u6570<\/p>\n
Problem 1:<\/b>\u00a0Give a finite set of positive integers T<\/span>. Let \\mathfrak{T}_n<\/span> be the collection of sequences (t_1,t_2,...,t_m)<\/span>, such that \\sum_{i=1}^m t_m=n<\/span>, and each t_i\\in T<\/span>. Let a_n=|\\mathfrak{T}_n|<\/span> for n\\ge1<\/span> and a_0=1<\/span>, and f(x)=\\sum_{n=0}^{\\infty}a_n x^n<\/span>. Find out what is f(x)<\/span> explicitly.<\/p>\n Solution<\/b>: It is not hard to note the recursive relation a_n=\\sum_{t\\in T} a_{n-t}<\/span> for n\\ge1<\/span>, if we set a_i=0<\/span> for negative i<\/span>. So that f(x)=1+\\sum_{t\\in T} x^t f(x)<\/span> and f(x)=1\/(1-\\sum_{t\\in T} x^t)<\/span>, which is a rational function.<\/p>\n Variantion 1:<\/b>\u00a0If T<\/span> is infinite, what would happen? Would f(x)<\/span> still be rational?<\/p>\n We first analyze simple cases. If T=\\mathbb{N}^+<\/span>, it is expected that f(x)=1+\\sum_{t=1}^{\\infty} x^t f(x)=1+f(x) x\/(1-x)<\/span>, so that f(x)=(1-x)\/(1-2x)=1+\\sum_{n=1}^{\\infty} 2^{n-1} x^n<\/span>. Indeed, in this case, counting the sequences amounts to divide a sequence of n<\/span> objects arbitrarily. You can choose to divide between the i<\/span>th and i+1<\/span>th for 1\\ge i\\ge n-1<\/span>, so in all 2^{n-1}<\/span> choices, justifying the expansion.<\/p>\n I think it is equivalent to f<\/span> being rational.<\/p>\n Theorem: \\mathbb{Q}_p(t)\\cap\\mathbb{Q}[[t]]=\\mathbb{Q}(t)<\/span><\/p>\n It is very interesting that this theorem is used for the rationality of \\zeta<\/span>-functions for algebraic varieties, which is part of the Weil conjectures.<\/p>\n 2013 \u5e74 7 \u6708 10 \u65e5<\/p>\n <\/p>","protected":false},"excerpt":{"rendered":"